Original questions in the style of Papers 1 and 2, each with a full worked mark scheme. Tick questions to build a printable test. Topic 3 at Higher Level →
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1AASL3-q00013.1 Three-dimensional geometryPaper 1Easy7 marks
The points \(A\) and \(B\) have coordinates \((-1,\,4,\,2)\) and \((3,\,-2,\,14)\) respectively.
(a)
Find the distance \(AB\).
(2)
(b)
Find the coordinates of \(M\), the midpoint of \([AB]\).
(2)
(c)
The point \(C\) has coordinates \((3,\,k,\,2)\), where \(k\in\mathbb{R}\). Given that \(CM=7\), find the possible values of \(k\).
2AASL3-q00023.1 Three-dimensional geometryPaper 1Easy7 marks
Diagram not to scale
A toy is made from a solid right circular cone joined to a solid hemisphere. The cone and the hemisphere have a common circular face of radius \(3\) cm, and the height of the cone is \(8\) cm, as shown in the diagram.
(a)
Find the volume of the toy. Give your answer in the form \(k\pi\text{ cm}^{3}\), where \(k\in\mathbb{Z}\).
4AASL3-q00043.1 Three-dimensional geometryPaper 1Easy6 marks
A cylindrical container with internal radius \(6\) cm stands on a horizontal table and contains water. Metal balls, each a sphere of radius \(1.5\) cm, are dropped into the container. Every ball sinks and is completely covered by the water, and no water overflows.
(a)
Find the exact volume of one ball.
(2)
(b)
Twenty balls are dropped into the container. Show that the water level rises by \(2.5\) cm.
(2)
(c)
The container is emptied of balls, refilled to its original level, and \(n\) balls are dropped in. Find the least value of \(n\) for which the water level rises by at least \(3.7\) cm.
19AASL3-q00053.1 Three-dimensional geometryPaper 1Medium8 marks
A solid hemisphere and a solid right circular cone both have base radius \(r\) cm. The total surface area of the hemisphere (including its flat face) is equal to the total surface area of the cone (including its base).
(a)
Show that the slant height of the cone is \(2r\) cm.
(2)
(b)
Find the exact height of the cone in terms of \(r\).
(1)
(c)
Show that \(\dfrac{\text{volume of the cone}}{\text{volume of the hemisphere}}=\dfrac{\sqrt{3}}{2}\).
(3)
(d)
Find the exact angle, in radians, between a slant edge of the cone and its base.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(2\pi r^{2}+\pi r^{2}=\pi r^{2}+\pi rl\)
M1
Hemisphere: curved 2πr² plus flat πr²; cone: base plus curved surface.
\(3\pi r^{2}-\pi r^{2}=\pi rl\) or \(2\pi r^{2}=\pi rl\)
22AASL3-q00173.2 Right-angled and non-right-angled trigonometryPaper 1Medium7 marks
The lengths of the sides of a triangle are \(x\) cm, \((x+1)\) cm and \((x+2)\) cm, where \(x\gt0\). The largest angle of the triangle is \(120^{\circ}\).
(a)
Show that \(2x^{2}-x-3=0\).
(3)
(b)
Find the value of \(x\).
(2)
(c)
Find the exact area of the triangle.
(2)
Show mark scheme
Working
Mark
Notes
(a)
the largest angle is opposite the longest side: \((x+2)^{2}=x^{2}+(x+1)^{2}-2x(x+1)\cos120^{\circ}\)
M1
\(\cos120^{\circ}=-\dfrac{1}{2}\), so \((x+2)^{2}=x^{2}+(x+1)^{2}+x(x+1)\)
23AASL3-q00243.3 Applications of trigonometryPaper 1Medium6 marks
A surveyor stands at point \(P\) on horizontal ground and measures the angle of elevation of the top of a vertical tower as \(45^{\circ}\). She then walks \(18\) m directly towards the base of the tower to point \(Q\), where the angle of elevation of the top of the tower is \(60^{\circ}\). The height of the tower is \(h\) m.
(a)
Show that \(h=27+9\sqrt{3}\).
(4)
(b)
Find the exact distance from \(Q\) to the base of the tower.
(2)
Show mark scheme
Working
Mark
Notes
(a)
distance from \(P\) to the base \(=\dfrac{h}{\tan45^{\circ}}=h\)
M1
distance from \(Q\) to the base \(=\dfrac{h}{\tan60^{\circ}}=\dfrac{h}{\sqrt{3}}\)
24AASL3-q00363.4 Radian measure, arcs and sectorsPaper 1Medium7 marks
The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(r\) cm, where \(A\hat{O}B=\dfrac{\pi}{3}\). The shaded region is the segment bounded by the chord \([AB]\) and the arc \(AB\).
(a)
Show that the area of the shaded region is \(\dfrac{r^{2}}{12}\left(2\pi-3\sqrt{3}\right)\text{ cm}^{2}\).
(3)
(b)
The area of the shaded region is \(\left(6\pi-9\sqrt{3}\right)\text{ cm}^{2}\). Find the value of \(r\).
(2)
(c)
Hence find the exact perimeter of the shaded region.
(2)
Show mark scheme
Working
Mark
Notes
(a)
area of sector \(=\dfrac{1}{2}r^{2}\left(\dfrac{\pi}{3}\right)=\dfrac{\pi r^{2}}{6}\)
M1
area of triangle \(=\dfrac{1}{2}r^{2}\sin\dfrac{\pi}{3}=\dfrac{\sqrt{3}}{4}r^{2}\)
Award A0 if \(r=-6\) is also given as a final answer.
(c)
arc \(=6\times\dfrac{\pi}{3}=2\pi\) and chord \(AB=6\) (triangle \(OAB\) is equilateral)
(M1)
FT from (b).
\(\left(2\pi+6\right)\) cm
A1
Answers: (b) \(r=6\) · (c) \(2\pi+6\) cm
25AASL3-q00383.4 Radian measure, arcs and sectorsPaper 1Medium6 marks
A sector of a circle has radius \(15\) cm and angle \(\theta\) radians. The two straight edges of the sector are joined, without overlap, to form the curved surface of a right circular cone with base radius \(9\) cm.
(a)
Find the exact value of \(\theta\).
(2)
(b)
Find the height of the cone.
(2)
(c)
Find the exact volume of the cone.
(2)
Show mark scheme
Working
Mark
Notes
(a)
arc length of the sector = circumference of the base: \(15\theta=2\pi(9)\)
(M1)
\(\theta=\dfrac{6\pi}{5}\)
A1
(b)
slant height \(=15\): \(h=\sqrt{15^{2}-9^{2}}\)
(M1)
Recognizing the radius of the sector becomes the slant height.
26AASL3-q00393.4 Radian measure, arcs and sectorsPaper 1Medium9 marks
Diagram not to scale
The diagram shows a circle with centre \(O\) and radius \(r\) cm. The point \(T\) is such that \(OT=2r\) cm. The lines \((TA)\) and \((TB)\) are tangents to the circle at \(A\) and \(B\). The shaded region is bounded by \([TA]\), \([TB]\) and the minor arc \(AB\).
(a)
Show that \(A\hat{O}T=\dfrac{\pi}{3}\).
(2)
(b)
Show that the area of the shaded region is \(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}\text{ cm}^{2}\).
(3)
(c)
Find an expression, in terms of \(r\), for the exact perimeter of the shaded region.
(2)
(d)
For one value of \(r\), the area of the shaded region in cm2 is numerically equal to its perimeter in cm. Show that this value is \(r=\dfrac{2\left(3\sqrt{3}+\pi\right)}{3\sqrt{3}-\pi}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(O\hat{A}T=\dfrac{\pi}{2}\) (tangent perpendicular to radius), so \(\cos A\hat{O}T=\dfrac{r}{2r}\)
M1
\(\cos A\hat{O}T=\dfrac{1}{2}\)
A1
\(A\hat{O}T=\dfrac{\pi}{3}\)
AG
(b)
area of kite \(OATB=2\times\dfrac{1}{2}(r)\left(\sqrt{3}\,r\right)=\sqrt{3}\,r^{2}\)
M1
\(AT=\sqrt{4r^{2}-r^{2}}=\sqrt{3}\,r\).
area of sector \(OAB=\dfrac{1}{2}r^{2}\left(\dfrac{2\pi}{3}\right)=\dfrac{\pi r^{2}}{3}\)
M1
\(\sqrt{3}\,r^{2}-\dfrac{\pi r^{2}}{3}\)
A1
Must see the subtraction.
\(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}\)
AG
(c)
\(TA=TB=\sqrt{3}\,r\) and arc \(AB=r\left(\dfrac{2\pi}{3}\right)\)
(M1)
Award for either correct length.
\(\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) cm
A1
Accept equivalent forms, e.g. \(\dfrac{2r\left(3\sqrt{3}+\pi\right)}{3}\).
(d)
\(\left(\sqrt{3}-\dfrac{\pi}{3}\right)r^{2}=\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) and dividing by \(r\;(r\neq0)\)
Answers: (c) \(\left(2\sqrt{3}+\dfrac{2\pi}{3}\right)r\) cm
27AASL3-q00453.5 The unit circle and the ambiguous casePaper 1Medium7 marks
The point \(P\left(-\dfrac{5}{13},\,\dfrac{12}{13}\right)\) lies on the unit circle with centre \(O\). The angle between the positive \(x\)-axis and \([OP]\), measured anticlockwise, is \(\theta\).
(a)
Write down the value of \(\tan\theta\).
(1)
(b)
Find the value of \(\sin2\theta\).
(2)
(c)
Find the value of \(\cos2\theta\).
(2)
(d)
The point \(Q\) on the unit circle is such that the angle between the positive \(x\)-axis and \([OQ]\), measured anticlockwise, is \(\theta+\pi\). Write down the coordinates of \(Q\).
28AASL3-q00483.5 The unit circle and the ambiguous casePaper 1Medium6 marks
The line \(L\) passes through the origin \(O\) and makes an angle of \(\dfrac{2\pi}{3}\) with the positive \(x\)-axis, measured anticlockwise.
(a)
Find the equation of \(L\), giving your answer in the form \(y=mx\), where \(m\) is exact.
(2)
(b)
The line \(L\) meets the unit circle \(x^{2}+y^{2}=1\) at the points \(A\) and \(B\), where \(A\) lies in the second quadrant. Find the exact coordinates of \(A\) and of \(B\).
(2)
(c)
The line \(M\) has equation \(y=x\). Find the acute angle between \(L\) and \(M\).
30AASL3-q00503.5 The unit circle and the ambiguous casePaper 1Medium9 marks
The point \(P(\cos\theta,\,\sin\theta)\) lies on the unit circle with centre \(O\), where \(0\lt\theta\lt\dfrac{\pi}{2}\). The point \(Q\) is the reflection of \(P\) in the \(y\)-axis and the point \(R\) is the reflection of \(P\) in \(O\).
Diagram not to scale
(a)
Write down the coordinates of \(Q\) and of \(R\) in terms of \(\theta\).
(2)
(b)
Show that the area of triangle \(PQR\) is \(2\sin\theta\cos\theta\).
(2)
(c)
The points \(Q\) and \(R\) can also be written as \(Q(\cos\alpha,\,\sin\alpha)\) and \(R(\cos\beta,\,\sin\beta)\), where \(0\lt\alpha\lt2\pi\) and \(0\lt\beta\lt2\pi\). Write down \(\alpha\) and \(\beta\) in terms of \(\theta\).
(2)
(d)
Given that the gradient of the line \((RP)\) is \(\dfrac12\), find the exact area of triangle \(PQR\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(Q(-\cos\theta,\,\sin\theta)\)
A1
\(R(-\cos\theta,\,-\sin\theta)\)
A1
(b)
\(PQ=2\cos\theta\) and \(QR=2\sin\theta\), with a right angle at \(Q\)
M1
area \(=\dfrac12(2\cos\theta)(2\sin\theta)\)
A1
\(=2\sin\theta\cos\theta\)
AG
(c)
\(\alpha=\pi-\theta\)
A1
\(\beta=\pi+\theta\)
A1
(d)
\((RP)\) passes through \(O\), so \(\tan\theta=\dfrac12\)
(M1)
line through the origin: \(y=x\tan\theta\)
\(\sin\theta=\dfrac{1}{\sqrt5}\) and \(\cos\theta=\dfrac{2}{\sqrt5}\)
(M1)
e.g. from a right-angled triangle with sides 1, 2, \(\sqrt5\) (\(\theta\) acute).
area \(=2\left(\dfrac{1}{\sqrt5}\right)\left(\dfrac{2}{\sqrt5}\right)=\dfrac45\)
A1
Or \(\dfrac12(2\cos\theta)(2\sin\theta)\) with their values.
32AASL3-q00603.6 Trigonometric identitiesPaper 1Medium9 marks
The diagram shows triangle \(ABC\) with \(AB=AC=6\) cm and \(B\hat{A}C=2\theta\), where \(\cos\theta=\dfrac{2}{3}\) and \(0\lt\theta\lt\dfrac{\pi}{2}\).
Diagram not to scale
(a)
Show that \(\sin2\theta=\dfrac{4\sqrt5}{9}\).
(3)
(b)
Hence find the exact area of triangle \(ABC\).
(2)
(c)
Find the exact value of \(\cos2\theta\).
(2)
(d)
Hence, or otherwise, find the exact length of \([BC]\).
35AASL3-q00663.7 Circular functions and their graphsPaper 1Medium7 marks
Let \(f(x)=a\sin\big(b(x+c)\big)+d\), where \(a,b\gt0\) and \(0\lt c\lt\dfrac{\pi}{2}\). The diagram shows the graph of \(y=f(x)\) for \(0\le x\le\pi\). The graph has a maximum point at \(\left(\dfrac{\pi}{6},\,5\right)\) and a minimum point at \(\left(\dfrac{2\pi}{3},\,-1\right)\).
(a)
Write down the value of \(a\) and the value of \(d\).
(2)
(b)
Find the value of \(b\).
(2)
(c)
Find the value of \(c\).
(2)
(d)
Write down the number of solutions of the equation \(f(x)=4\) for \(0\le x\le\pi\).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(a=3\)
A1
\(d=2\)
A1
(b)
period \(=2\left(\dfrac{2\pi}{3}-\dfrac{\pi}{6}\right)=\pi\)
(M1)
\(b=\dfrac{2\pi}{\pi}=2\)
A1
(c)
\(2\left(\dfrac{\pi}{6}+c\right)=\dfrac{\pi}{2}\)
(M1)
Or substituting \(\left(\frac{\pi}{6},5\right)\): \(3\sin\big(2(\frac{\pi}{6}+c)\big)+2=5\). FT from (a), (b).
37AASL3-q00703.7 Circular functions and their graphsPaper 1Medium9 marks
Let \(f(x)=2\cos2x+1\), for \(0\le x\le\pi\).
(a)
Find the \(x\)-intercepts of the graph of \(f\).
(3)
(b)
Sketch the graph of \(y=f(x)\), showing the coordinates of the end points and of the minimum point.
(3)
(c)
Hence find the set of values of \(x\) for which \(f(x)\gt2\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\cos2x=-\dfrac12\), with \(0\le2x\le2\pi\)
M1
\(2x=\dfrac{2\pi}{3},\ \dfrac{4\pi}{3}\)
(A1)
\(x=\dfrac{\pi}{3},\ \dfrac{2\pi}{3}\)
A1
(b)
end points \((0,\,3)\) and \((\pi,\,3)\)
A1
minimum point \(\left(\dfrac{\pi}{2},\,-1\right)\)
A1
correct shape: smooth curve symmetric about \(x=\dfrac{\pi}{2}\), crossing the \(x\)-axis at their intercepts
A1
(c)
\(\cos2x=\dfrac12\)
M1
\(x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}\)
A1
\(0\le x\lt\dfrac{\pi}{6}\) or \(\dfrac{5\pi}{6}\lt x\le\pi\)
A1
Accept strict or non-strict inequality at 0 and π; endpoints π/6 and 5π/6 must be excluded.
Answers: (a) \(\frac{\pi}{3},\frac{2\pi}{3}\) · (c) \(0\le x\lt\frac{\pi}{6}\) or \(\frac{5\pi}{6}\lt x\le\pi\)
38AASL3-q00713.7 Circular functions and their graphsPaper 1Medium10 marks
The graph of \(y=\sin x\) is transformed onto the graph of \(y=g(x)\) by the following sequence of transformations:
a vertical stretch with scale factor \(3\);
a horizontal stretch with scale factor \(\dfrac12\);
a translation by the vector \(\begin{pmatrix}\frac{\pi}{6}\\-1\end{pmatrix}\).
(a)
Find \(g(x)\), giving your answer in the form \(g(x)=a\sin\big(b(x+c)\big)+d\).
(3)
(b)
The point \(\left(\dfrac{\pi}{2},\,1\right)\) on the graph of \(y=\sin x\) is mapped to the point \(M\) on the graph of \(y=g(x)\). Find the coordinates of \(M\).
(2)
(c)
Find the exact value of \(g(0)\).
(2)
(d)
Solve the equation \(g(x)=\dfrac12\), for \(0\le x\le\pi\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(a=3\)
A1
\(b=2\)
A1
\(c=-\dfrac{\pi}{6}\) and \(d=-1\), so \(g(x)=3\sin\left(2\left(x-\dfrac{\pi}{6}\right)\right)-1\)
A1
Award A0 for \(3\sin\left(2x-\frac{\pi}{6}\right)-1\).
\(A\hat{C}B\) may be acute (\(C_{1}\)) or obtuse, \(C_{2}=180^{\circ}-C_{1}\), since both have sine \(\dfrac{5}{6}\)
R1
since \(\dfrac{5}{6}\gt\dfrac{1}{2}=\sin30^{\circ}\), \(C_{1}\gt30^{\circ}\), so \(C_{2}+30^{\circ}\lt180^{\circ}\) and the obtuse case gives a valid triangle
\(6^{2}=10^{2}+x^{2}-2(10)x\cos30^{\circ}\), where \(x=AC\)
M1
\(x^{2}-10\sqrt{3}\,x+64=0\)
A1
\(x=\dfrac{10\sqrt{3}\pm\sqrt{300-256}}{2}\)
M1
Correct substitution into the quadratic formula, or completing the square.
\(AC=5\sqrt{3}+\sqrt{11}\) or \(AC=5\sqrt{3}-\sqrt{11}\)
A1
Both values required.
Answers: (c) \(AC=5\sqrt{3}\pm\sqrt{11}\) cm
42AASL3-q00543.5 The unit circle and the ambiguous casePaper 1Hard13 marks
In triangle \(ABC\), \(AB=12\) cm, \(B\hat{A}C=30^{\circ}\), \(BC=x\) cm and \(AC=y\) cm.
(a)
Show that \(y^{2}-12\sqrt3\,y+144-x^{2}=0\).
(2)
(b)
By considering this equation as a quadratic in \(y\), find the set of values of \(x\) for which two different triangles \(ABC\) can be drawn.
(4)
(c)
For a particular value of \(x\), two different triangles can be drawn and the length of \([AC]\) in one triangle is twice the length of \([AC]\) in the other. Find this value of \(x\).
(4)
(d)
For the value of \(x\) found in part (c), find the two possible sizes of \(A\hat{C}B\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(x^{2}=12^{2}+y^{2}-2(12)(y)\cos30^{\circ}\)
M1
\(\cos30^{\circ}=\dfrac{\sqrt3}{2}\), so \(x^{2}=144+y^{2}-12\sqrt3\,y\)
A1
\(y^{2}-12\sqrt3\,y+144-x^{2}=0\)
AG
(b)
two distinct roots: \(\left(12\sqrt3\right)^{2}-4\left(144-x^{2}\right)\gt0\)
M1
\(4x^{2}\gt144\), so \(x\gt6\)
A1
since \(x\gt0\)
both roots must be positive: their sum is \(12\sqrt3\gt0\), so their product \(144-x^{2}\gt0\)
R1
\(6\lt x\lt12\)
A1
(c)
roots \(k\) and \(2k\): \(3k=12\sqrt3\)
M1
\(k=4\sqrt3\)
A1
so \(AC=4\sqrt3\) or \(8\sqrt3\)
product of roots \(2k^{2}=144-x^{2}\), i.e. \(96=144-x^{2}\)
M1
Or substitute \(y=4\sqrt3\) into the equation in (a).
\(x=4\sqrt3\)
A1
Accept \(\sqrt{48}\). Award A0 if \(-4\sqrt3\) is also given.
(d)
valid method for one triangle, e.g. \(AC=BC=4\sqrt3\), so the triangle is isosceles with \(A\hat{B}C=30^{\circ}\); or \(AB^{2}+BC^{2}=144+48=192=AC^{2}\)
(M1)
Or sine rule: \(\sin A\hat{C}B=\dfrac{12\sin30^{\circ}}{4\sqrt3}=\dfrac{\sqrt3}{2}\).
\(\cos x\lt0\) and \(\sin x\gt0\) for \(\dfrac{\pi}{2}\lt x\lt\pi\), so \(\cos x-\sin x\lt0\)
R1
Alternative R1: \(\sin x+\cos x\gt0\) with \(\cos x\lt0\) gives \(\dfrac{\pi}{2}\lt x\lt\dfrac{3\pi}{4}\), so \(\pi\lt2x\lt\dfrac{3\pi}{2}\) and \(\cos2x\lt0\).
\(\cos x-\sin x=-\dfrac{\sqrt{7}}{2}\)
A1
In the alternative method, award for \(\cos2x=\pm\dfrac{\sqrt{7}}{4}\).
49AASL3-q00103.1 Three-dimensional geometryPaper 2Easy9 marks
A fuel storage tank is in the shape of a right circular cylinder with a hemisphere attached at each end. The cylinder has radius \(3.8\) m and length \(13.8\) m, and each hemisphere has radius \(3.8\) m.
(a)
Find the volume of the tank.
(3)
(b)
Find the external surface area of the tank.
(3)
(c)
The outside of the tank is to be painted with one coat of paint. One litre of paint covers \(12\text{ m}^{2}\) and the paint is sold in \(5\)-litre tins. Find the least number of tins needed.
(3)
Show mark scheme
Working
Mark
Notes
(a)
cylinder: \(\pi(3.8)^{2}(13.8)\;(=626.03)\)
(M1)
two hemispheres form a sphere: \(\dfrac{4}{3}\pi(3.8)^{3}\;(=229.85)\)
\(87.0636\ldots\) Accept \(87.4\) from using \(13.2\).
Answers: (a) \(13.2\) cm · (b) \(87.1\text{ cm}^{2}\)
51AASL3-q00203.2 Right-angled and non-right-angled trigonometryPaper 2Easy7 marks
Building regulations state that a wheelchair ramp must rise no more than \(1\) m vertically for every \(12\) m horizontally. A straight ramp is to be built from the ground to a doorway \(0.75\) m above the ground.
(a)
Write down the least horizontal distance the ramp must cover.
(1)
(b)
Find the greatest angle that the ramp may make with the horizontal.
(2)
(c)
Find the length of the sloping surface of the ramp in part (a).
(2)
(d)
A builder proposes a straight ramp whose sloping surface is \(8.5\) m long, from the ground to the doorway. Determine whether this ramp meets the regulations.
\(5.06209\ldots^{\circ}\). Or \(0.0884\) radians. Or horizontal distance \(\sqrt{8.5^{2}-0.75^{2}}=8.47\lt9\).
\(5.06^{\circ}\gt4.76^{\circ}\), so the ramp is too steep and does not meet the regulations
R1
Do not award R1 without a correct comparison. Award (M1)R0 for a correct angle with an incorrect conclusion.
Answers: (a) \(9\) m · (b) \(4.76^{\circ}\) · (c) \(9.03\) m · (d) No (\(5.06^{\circ}\gt4.76^{\circ}\))
52AASL3-q00253.3 Applications of trigonometryPaper 2Easy5 marks
A vertical flagpole \([AB]\) of height \(7\) m stands on horizontal ground, with its base at \(B\). From a point \(C\) on the ground the angle of elevation of the top of the flagpole, \(A\), is \(32^{\circ}\). From a point \(D\) on \([CB]\) the angle of elevation of \(A\) is \(55^{\circ}\).
(a)
Find the distance \(BC\).
(2)
(b)
Find the distance \(CD\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan32^{\circ}=\dfrac{7}{BC}\)
(M1)
\(BC=11.2\) m
A1
\(11.2023\ldots\)
(b)
\(\tan55^{\circ}=\dfrac{7}{BD}\)
(M1)
\(BD=4.90\) m
(A1)
\(4.90145\ldots\)
\(CD=11.202\ldots-4.901\ldots=6.30\) m
A1
\(6.30088\ldots\) Accept \(6.3\). FT from (a).
Answers: (a) \(11.2\) m · (b) \(6.30\) m
53AASL3-q00293.3 Applications of trigonometryPaper 2Easy5 marks
The banks of a river are straight and parallel. Two points \(P\) and \(Q\) are \(30\) m apart on one bank. A tree \(T\) stands on the opposite bank. It is measured that \(T\hat{P}Q=75^{\circ}\) and \(T\hat{Q}P=50^{\circ}\).
Perpendicular distance from \(T\) to the line \((PQ)\). FT from (b).
\(27.1\) m
A1
\(27.0990\ldots\)
Answers: (a) \(55^{\circ}\) · (b) \(35.4\) m · (c) \(27.1\) m
54AASL3-q00413.4 Radian measure, arcs and sectorsPaper 2Easy7 marks
Diagram not to scale
A windscreen wiper rotates about the point \(O\) through an angle of \(2.2\) radians. The rubber blade lies along the arm, from \(12\) cm to \(54\) cm from \(O\). The shaded region shows the area of the windscreen wiped by the blade.
(a)
Find the area wiped by the blade.
(2)
(b)
Find the perimeter of the wiped region.
(3)
(c)
The wiper takes \(0.75\) seconds to rotate through the \(2.2\) radians. Find the average speed of the tip of the blade, in cm s−1.
the other angle with this sine is \(180^{\circ}-27.9^{\circ}=152.1^{\circ}\)
A1
\(152.1^{\circ}+40^{\circ}\gt 180^{\circ}\), so this angle is impossible
R1
Accept: the side opposite the given angle (11 cm) is longer than \(PQ\), so \(P\hat{R}Q\) must be acute.
(c)
\(P\hat{Q}R=180^{\circ}-40^{\circ}-27.87^{\circ}=112.1^{\circ}\) and \(\dfrac{PR}{\sin112.1^{\circ}}=\dfrac{11}{\sin40^{\circ}}\)
(M1)
\(15.9\) cm
A1
\(15.8523\ldots\)
Answers: (a) \(27.9^{\circ}\) · (c) 15.9 cm
56AASL3-q00063.1 Three-dimensional geometryPaper 2Medium8 marks
A solid metal sphere has volume \(500\text{ cm}^{3}\).
(a)
Find the radius of the sphere.
(2)
(b)
Find the surface area of the sphere.
(2)
(c)
The sphere is melted down and all of the metal is recast into a solid right circular cone. The radius of the base of the cone is equal to the radius, \(r\) cm, of the sphere. Show that the height of the cone is \(4r\) cm.
(2)
(d)
Find the curved surface area of the cone.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{4}{3}\pi r^{3}=500\)
(M1)
\(r=4.92\) cm
A1
\(4.92372\ldots\)
(b)
\(4\pi(4.92372\ldots)^{2}\)
(M1)
FT from their \(r\).
\(305\text{ cm}^{2}\)
A1
\(304.647\ldots\) Accept \(304\)–\(305\).
(c)
\(\dfrac{1}{3}\pi r^{2}h=\dfrac{4}{3}\pi r^{3}\)
M1
Equating the two volumes in terms of r. Award M0 for working only with 500 and a numerical r.
\(r^{2}h=4r^{3}\)
A1
\(h=4r\)
AG
(d)
slant height \(l=\sqrt{r^{2}+(4r)^{2}}=r\sqrt{17}\;(=20.3)\) and \(\pi rl\)
57AASL3-q00073.1 Three-dimensional geometryPaper 2Medium8 marks
Diagram not to scale
The diagram shows a right pyramid \(VABCD\) with a square base of side \(10\) cm. The point \(M\) is the centre of the base and \(V\) is vertically above \(M\), with \(VM=12\) cm.
(a)
Find the volume of the pyramid.
(2)
(b)
Find the angle between the edge \(VA\) and the base \(ABCD\).
58AASL3-q00093.1 Three-dimensional geometryPaper 2Medium8 marks
Diagram not to scale
A solid is made from a right circular cylinder of radius \(r\) cm and height \(3r\) cm, with a solid right circular cone of base radius \(r\) cm and height \(2r\) cm fixed to its top face, as shown in the diagram.
(a)
Show that the volume of the solid is \(\dfrac{11}{3}\pi r^{3}\text{ cm}^{3}\).
(2)
(b)
The volume of the solid is \(2000\text{ cm}^{3}\). Find the value of \(r\).
(2)
(c)
Show that the slant height of the cone is \(\sqrt{5}\,r\) cm.
(1)
(d)
Find the total surface area of the solid.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\pi r^{2}(3r)+\dfrac{1}{3}\pi r^{2}(2r)\)
M1
Sum of the volumes of the cylinder and the cone, both in terms of r.
\(3\pi r^{3}+\dfrac{2}{3}\pi r^{3}\)
A1
\(\dfrac{11}{3}\pi r^{3}\)
AG
(b)
\(\dfrac{11}{3}\pi r^{3}=2000\)
(M1)
\(r=5.58\)
A1
\(5.57874\ldots\)
(c)
\(l=\sqrt{r^{2}+(2r)^{2}}=\sqrt{5r^{2}}\)
A1
Must see the Pythagoras step.
\(l=\sqrt{5}\,r\)
AG
(d)
cylinder: base \(\pi r^{2}\) and curved surface \(2\pi r(3r)=6\pi r^{2}\)
(M1)
Award M0 if the top face of the cylinder (inside the solid) is included.
cone: curved surface \(\pi r(\sqrt{5}\,r)\); total \(=(7+\sqrt{5})\pi r^{2}\)
(M1)
\(903\text{ cm}^{2}\)
A1
\(903.045\ldots\) FT from (b). Accept \(902\)–\(904\).
59AASL3-q00113.1 Three-dimensional geometryPaper 2Medium9 marks
Diagram not to scale
The diagram shows a right pyramid \(VABCD\) with a horizontal rectangular base \(ABCD\), where \(AB=16\) cm and \(BC=10\) cm. The point \(M\) is the point of intersection of the diagonals of the base, and \(V\) is vertically above \(M\) with \(VM=9\) cm.
(a)
Find the length of the edge \([VA]\).
(2)
(b)
The point \(N\) is the midpoint of \([BC]\). Find the angle between the line \([VN]\) and the base \(ABCD\).
(2)
(c)
Find the size of the angle \(A\hat{V}C\).
(2)
(d)
Find the total surface area of the pyramid.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(AM=\dfrac{1}{2}\sqrt{16^{2}+10^{2}}=\sqrt{89}\;(=9.43)\) and \(VA=\sqrt{89+9^{2}}\)
(M1)
\(VA=\sqrt{170}=13.0\) cm
A1
\(13.0384\ldots\)
(b)
\(MN=8\) and \(\tan V\hat{N}M=\dfrac{9}{8}\)
(M1)
Or \(VN=\sqrt{145}\) and \(\sin V\hat{N}M=\dfrac{9}{\sqrt{145}}\). Must identify angle \(V\hat{N}M\); award M0 for \(MN=5\).
\(48.4^{\circ}\)
A1
\(48.3664\ldots^{\circ}\); \(0.844\) radians
(c)
\(A\hat{V}C=2\times A\hat{V}M\), where \(\tan A\hat{V}M=\dfrac{\sqrt{89}}{9}\)
(M1)
Or \(180^{\circ}-2\times V\hat{A}M\) with \(\tan V\hat{A}M=\dfrac{9}{\sqrt{89}}\), or the cosine rule in triangle \(AVC\) with \(AC=2\sqrt{89}\).
\(92.7^{\circ}\)
A1
\(92.6972\ldots^{\circ}\); \(1.62\) radians
(d)
heights of the triangular faces: \(\sqrt{9^{2}+5^{2}}=\sqrt{106}\) (faces on \(AB\), \(CD\)) and \(\sqrt{9^{2}+8^{2}}=\sqrt{145}\) (faces on \(BC\), \(DA\))
60AASL3-q00133.1 Three-dimensional geometryPaper 2Medium9 marks
A vertical pole \([OT]\) of height \(12\) m stands on horizontal ground. A coordinate system is set up with \(O\) at the origin, the \(x\)-axis pointing east, the \(y\)-axis pointing north and the \(z\)-axis vertically upwards; units are metres. The ground is the plane \(z=0\). Two straight ropes join the top of the pole, \(T\), to the points \(A(9,\,-4,\,0)\) and \(B(-3,\,8,\,0)\) on the ground.
(a)
Write down the coordinates of \(T\).
(1)
(b)
Find the length of the rope \([TA]\).
(2)
(c)
Find the angle between the rope \([TA]\) and the ground.
(2)
(d)
Find the coordinates of \(M\), the midpoint of \([AB]\).
(1)
(e)
Find the angle of elevation of \(T\) from \(M\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(T(0,\,0,\,12)\)
A1
(b)
\(\sqrt{9^{2}+(-4)^{2}+(-12)^{2}}\)
(M1)
\(\sqrt{241}=15.5\) m
A1
\(15.5241\ldots\)
(c)
\(OA=\sqrt{97}\) and \(\tan T\hat{A}O=\dfrac{12}{\sqrt{97}}\)
(M1)
Or \(\sin T\hat{A}O=\dfrac{12}{\sqrt{241}}\).
\(50.6^{\circ}\)
A1
\(50.6229\ldots^{\circ}\); \(0.884\) radians
(d)
\(M(3,\,2,\,0)\)
A1
(e)
\(OM=\sqrt{3^{2}+2^{2}}=\sqrt{13}\)
(A1)
FT from (d).
\(\tan T\hat{M}O=\dfrac{12}{\sqrt{13}}\)
(M1)
Or \(TM=\sqrt{157}=12.5\) and \(\sin T\hat{M}O=\dfrac{12}{\sqrt{157}}\).
Answers: (b) \(48.6^{\circ}\) or \(131^{\circ}\) · (c) \(16.4\) cm · (d) \(34.4\) cm
63AASL3-q00263.3 Applications of trigonometryPaper 2Medium6 marks
Diagram not to scale
A vertical cliff \([TF]\) is \(45\) m high, where \(F\) is at sea level. A boat sails in a straight line directly towards \(F\). At point \(A\) the angle of depression of the boat from \(T\) is \(12^{\circ}\). Two minutes later the boat is at point \(B\) and the angle of depression is \(20^{\circ}\).
(a)
Find the distance \(FA\).
(2)
(b)
Find the distance \(AB\).
(2)
(c)
Find the average speed of the boat between \(A\) and \(B\), giving your answer in km h−1.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan12^{\circ}=\dfrac{45}{FA}\)
(M1)
Angle of elevation of \(T\) from \(A\) is \(12^{\circ}\) (alternate angles).
\(FA=212\) m
A1
\(211.708\ldots\)
(b)
\(FB=\dfrac{45}{\tan20^{\circ}}\;(=123.636\ldots)\) and subtracting
(M1)
\(AB=88.1\) m
A1
\(88.0719\ldots\) FT from (a).
(c)
\(\dfrac{88.07\ldots}{120}=0.734\text{ m s}^{-1}\) or \(\dfrac{0.08807\ldots}{\tfrac{2}{60}}\)
(M1)
Correct conversion of both units.
\(2.64\text{ km h}^{-1}\)
A1
\(2.64215\ldots\) FT from (b).
Answers: (a) \(212\) m · (b) \(88.1\) m · (c) \(2.64\text{ km h}^{-1}\)
64AASL3-q00273.3 Applications of trigonometryPaper 2Medium8 marks
Two points \(A\) and \(B\) lie on horizontal ground, \(120\) m apart. A vertical tower \([FT]\) has its base \(F\) on the line segment \([AB]\). The angle of elevation of the top of the tower, \(T\), is \(28^{\circ}\) from \(A\) and \(41^{\circ}\) from \(B\).
(a)
Write down the size of \(A\hat{T}B\).
(1)
(b)
Find \(AT\).
(2)
(c)
Find the height of the tower.
(2)
(d)
An observer walks from \(A\) directly towards the tower and stops at the midpoint of \([AB]\). Find the angle of elevation of \(T\) from this point.
Or \(\dfrac{h}{\tan28^{\circ}}+\dfrac{h}{\tan41^{\circ}}=120\).
\(39.6\) m
A1
\(39.5896\ldots\)
(d)
\(AF=\dfrac{39.58\ldots}{\tan28^{\circ}}=74.46\ldots\) or \(84.32\ldots\cos28^{\circ}\)
(M1)
horizontal distance from the midpoint to \(F\) \(=74.46\ldots-60=14.46\ldots\) and \(\tan\theta=\dfrac{39.58\ldots}{14.46\ldots}\)
(M1)
\(69.9^{\circ}\)
A1
\(69.9388\ldots^{\circ}\) FT from (c).
Answers: (a) \(111^{\circ}\) · (b) \(84.3\) m · (c) \(39.6\) m · (d) \(69.9^{\circ}\)
65AASL3-q00303.3 Applications of trigonometryPaper 2Medium11 marks
A vertical radio mast \([MT]\) of height \(40\) m stands on horizontal ground with its base at \(M\). Point \(A\) is \(30\) m from \(M\) on a bearing of \(030^{\circ}\), and point \(B\) is \(25\) m from \(M\) on a bearing of \(140^{\circ}\).
(a)
Find the angle of elevation of \(T\) from \(A\).
(2)
(b)
Find the distance \(AB\).
(3)
(c)
Find the bearing of \(B\) from \(A\).
(3)
(d)
A technician walks in a straight line from \(A\) to \(B\). Find the greatest angle of elevation of \(T\) from the technician during this walk.
66AASL3-q00403.4 Radian measure, arcs and sectorsPaper 2Medium7 marks
The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(r\) cm. The angle \(A\hat{O}B\) is \(\theta\) radians. The length of the arc \(AB\) is \(10\) cm and the area of the sector is \(35\text{ cm}^{2}\).
(a)
Find the value of \(r\) and the value of \(\theta\).
(4)
(b)
Find the area of the shaded segment.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(r\theta=10\) and \(\dfrac{1}{2}r^{2}\theta=35\)
(M1)
\(\dfrac{1}{2}r(r\theta)=35\Rightarrow5r=35\)
(M1)
Any valid method of eliminating one variable.
\(r=7\)
A1
\(\theta=\dfrac{10}{7}\;(=1.43)\)
A1
\(1.42857\ldots\)
(b)
area of triangle \(OAB=\dfrac{1}{2}(7)^{2}\sin\dfrac{10}{7}\;(=24.25)\)
(M1)
GDC must be in radian mode. Award M0 for \(\sin1.43^{\circ}\).
67AASL3-q00423.4 Radian measure, arcs and sectorsPaper 2Medium8 marks
A pendulum consists of a small bob on a light string of length \(80\) cm, attached to a fixed point \(O\). The bob swings in a vertical plane between two extreme positions \(A\) and \(B\). The angle \(A\hat{O}B\) is \(0.6\) radians, and the string is vertical when the bob is midway between \(A\) and \(B\).
(a)
Find the length of the arc travelled by the bob in moving from \(A\) to \(B\).
(2)
(b)
Write down \(A\hat{O}B\) in degrees.
(1)
(c)
Find the straight-line distance \(AB\).
(2)
(d)
Find the height of \(A\) above the lowest point of the bob’s path.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(s=80\times0.6\)
(M1)
\(48\) cm
A1
(b)
\(34.4^{\circ}\)
A1
\(34.3774\ldots^{\circ}\)
(c)
\(AB=2\times80\sin0.3\)
(M1)
Or \(AB^{2}=80^{2}+80^{2}-2(80)(80)\cos0.6\). GDC in radian mode.
\(47.3\) cm
A1
\(47.2832\ldots\)
(d)
angle between \([OA]\) and the vertical is \(0.3\) radians
(M1)
vertical depth of \(A\) below \(O\) \(=80\cos0.3\;(=76.43)\); height \(=80-80\cos0.3\)
(M1)
\(3.57\) cm
A1
\(3.57308\ldots\)
Answers: (a) \(48\) cm · (b) \(34.4^{\circ}\) · (c) \(47.3\) cm · (d) \(3.57\) cm
68AASL3-q00523.5 The unit circle and the ambiguous casePaper 2Medium8 marks
A straight path starts at point \(A\). A post \(B\) is \(12\) m from \(A\), and \([AB]\) makes an angle of \(38^{\circ}\) with the path. A goat is tied to the post with a rope of length \(8.5\) m. When the rope is taut, the goat can reach the path at exactly two points, \(C_{1}\) and \(C_{2}\), as shown.
Diagram not to scale
(a)
Find the two possible sizes of the angle \(A\hat{C}B\), where \(C\) is a point on the path with \(BC=8.5\) m.
(3)
(b)
Find the distance \(C_{1}C_{2}\).
(3)
(c)
Find the least length of rope for which the goat can just reach the path.
\(119.637\ldots^{\circ}\). Accept 120°. FT \(180^{\circ}-\) their acute angle.
(b)
triangle \(BC_{1}C_{2}\) is isosceles with base angles \(60.36\ldots^{\circ}\), or \(AC_{1}=5.25\) and \(AC_{2}=13.7\) (cosine rule or sine rule)
(M1)
\(C_{1}C_{2}=2(8.5)\cos60.36\ldots^{\circ}\), or \(13.6595\ldots-5.25275\ldots\)
(M1)
\(8.41\) m
A1
\(8.40675\ldots\)
(c)
perpendicular distance from \(B\) to the path, \(12\sin38^{\circ}\)
(M1)
\(7.39\) m
A1
\(7.38793\ldots\)
Answers: (a) \(60.4^{\circ}\) and \(119.6^{\circ}\) · (b) 8.41 m · (c) 7.39 m
69AASL3-q00653.6 Trigonometric identitiesPaper 2Medium10 marks
The diagram shows a semicircle with centre \(O\) and radius \(10\) cm. The rectangle \(PQRS\) has \(P\) and \(Q\) on the diameter and \(R\) and \(S\) on the arc, and \(O\) is the midpoint of \([PQ]\). The angle \(Q\hat{O}R=\theta\), where \(0\lt\theta\lt\dfrac{\pi}{2}\).
Diagram not to scale
(a)
Show that the area of the rectangle is \(100\sin2\theta\text{ cm}^{2}\).
(3)
(b)
Write down the maximum area of the rectangle and the value of \(\theta\) at which it occurs.
(2)
(c)
Find the values of \(\theta\) for which the area of the rectangle is \(80\text{ cm}^{2}\).
(3)
(d)
Find the maximum perimeter of the rectangle.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(QR=10\sin\theta\) and \(OQ=10\cos\theta\)
A1
area \(=2(10\cos\theta)(10\sin\theta)=100(2\sin\theta\cos\theta)\)
M1
using \(\sin2\theta=2\sin\theta\cos\theta\)
A1
area \(=100\sin2\theta\)
AG
(b)
\(100\text{ cm}^{2}\)
A1
\(\theta=\dfrac{\pi}{4}\)
A1
Accept 0.785.
(c)
\(100\sin2\theta=80\), e.g. graph of \(y=100\sin2\theta\) with \(y=80\)
(M1)
\(\theta=0.464\)
A1
\(0.463647\ldots\)
\(\theta=1.11\)
A1
\(1.10714\ldots\)
(d)
perimeter \(=40\cos\theta+20\sin\theta\), maximum from GDC
(M1)
maximum at \(\theta=0.464\)
\(44.7\) cm
A1
\(44.7213\ldots=20\sqrt5\)
Answers: (b) 100 cm² at \(\theta=\frac{\pi}{4}\) · (c) 0.464, 1.11 · (d) 44.7 cm
70AASL3-q00743.7 Circular functions and their graphsPaper 2Medium8 marks
The depth of water, \(d\) metres, at the entrance to a harbour \(t\) hours after midnight is modelled by \(d(t)=2.6\cos\big(q(t-3)\big)+4.2\), for \(0\le t\le24\), where \(q\gt0\). The time between consecutive high tides is \(12.4\) hours.
(a)
Write down the depth of water at high tide.
(1)
(b)
Show that \(q=\dfrac{5\pi}{31}\).
(2)
(c)
Find the time of the first low tide after midnight, giving your answer to the nearest minute in 24-hour clock format.
(2)
(d)
A ship can enter the harbour only when the depth of water is at least \(3\) m. Find the total length of time between midnight and noon during which the ship can enter the harbour.
71AASL3-q00753.7 Circular functions and their graphsPaper 2Medium9 marks
The blood pressure, \(P\) mmHg, of a patient at rest is modelled by \(P(t)=98+22\sin(2.5\pi t)\), where \(t\) is the time in seconds, \(t\ge0\). Each period of \(P\) corresponds to one heartbeat.
(a)
Write down the maximum and the minimum blood pressure.
(2)
(b)
Find the number of heartbeats per minute.
(2)
(c)
Find the blood pressure when \(t=1.1\).
(2)
(d)
During exercise, the patient’s maximum and minimum blood pressures are each \(20\,\%\) higher than at rest, and the heart beats \(96\) times per minute. The blood pressure is then modelled by \(Q(t)=p+q\sin(kt)\), where \(p,q,k\gt0\). Find the value of \(p\), the value of \(q\) and the exact value of \(k\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
maximum \(120\) mmHg
A1
minimum \(76\) mmHg
A1
(b)
period \(=\dfrac{2\pi}{2.5\pi}=0.8\) s
(M1)
\(\dfrac{60}{0.8}=75\)
A1
(c)
\(98+22\sin(2.75\pi)\)
(M1)
\(114\) mmHg
A1
\(113.556\ldots\)
(d)
new maximum \(144\), new minimum \(91.2\); \(p=\dfrac{144+91.2}{2}=117.6\)
A1
\(q=\dfrac{144-91.2}{2}=26.4\)
A1
FT their (a).
period \(\dfrac{60}{96}=0.625\) s, so \(k=\dfrac{2\pi}{0.625}=3.2\pi\)
A1
Accept \(\dfrac{16\pi}{5}\). Award A0 for 10.1 only (exact value required).
72AASL3-q00763.7 Circular functions and their graphsPaper 2Medium8 marks
The mean monthly temperature, \(T\) °C, in a city is modelled using sinusoidal regression on a GDC. The calculator gives
\(T(x)=6.2\sin(0.524x-2.09)+14.8\),
where \(x\) is the month number (\(x=1\) for January, \(x=2\) for February, and so on), for \(1\le x\le12\).
(a)
Write \(T(x)\) in the form \(a\sin\big(b(x+c)\big)+d\), giving the value of \(c\) to three significant figures.
(2)
(b)
Find the period of \(T\) and comment on its value in this context.
(2)
(c)
Find the maximum mean monthly temperature predicted by the model, and the month in which it occurs.
(2)
(d)
Find the values of \(x\) for which \(T(x)=18\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.524\left(x-\dfrac{2.09}{0.524}\right)\)
(M1)
\(T(x)=6.2\sin\big(0.524(x-3.99)\big)+14.8\), so \(c=-3.99\)
A1
\(-3.98854\ldots\)
(b)
\(\dfrac{2\pi}{0.524}=12.0\)
A1
\(11.9908\ldots\)
approximately 12 months: the temperatures repeat every year
R1
(c)
\(21.0\) °C
A1
July
A1
maximum at \(x=6.99\) \((6.98625\ldots)\)
(d)
graph of \(T\) with \(y=18\), or \(\sin(0.524x-2.09)=\dfrac{3.2}{6.2}\)
(M1)
\(x=5.02\) and \(x=8.95\)
A1
\(5.02352\ldots,\ 8.94898\ldots\)
Answers: (a) \(c=-3.99\) · (b) 12.0 months · (c) 21.0 °C in July · (d) 5.02, 8.95
73AASL3-q00773.7 Circular functions and their graphsPaper 2Medium10 marks
The number of hours of daylight, \(D\), in a town on day \(t\) of the year is modelled by
\(D(t)=3.1\cos\left(\dfrac{2\pi}{365}(t-172)\right)+12.2\), for \(1\le t\le365,\ t\in\mathbb{Z}\),
where \(t=1\) is 1 January.
(a)
Write down the maximum number of hours of daylight and the value of \(t\) at which it occurs.
(2)
(b)
Find the number of hours of daylight on day 50.
(2)
(c)
Find the number of days in the year on which there are more than \(14\) hours of daylight.
(3)
(d)
In a second town, the longest day has \(16.8\) hours of daylight and the shortest day has \(7.6\) hours. These occur on the same days of the year as in the first town, and the second town is modelled by \(E(t)=a\cos\left(\dfrac{2\pi}{365}(t-172)\right)+d\). Find the number of days in the year on which the second town has more hours of daylight than the first town.
for \(x\gt8\), \(\dfrac{x}{2}\gt4\), so the two sides cannot be equal
R1
(c)
one further solution for \(2\pi\lt x\le8\), at \(x=6.63\)
(M1)
\(6.62810\ldots\)
\(5\)
A1
Answers: (a) 0.879, 2.39, 3.78, 5.85 · (c) 5
77AASL3-q00083.1 Three-dimensional geometryPaper 2Hard9 marks
Diagram not to scale
The diagram shows a cuboid \(OABCDEFG\) in which \(OA=8\) cm, \(OC=6\) cm and \(OD=4\) cm. Coordinates are defined so that \(O\) is the origin and \(A\), \(C\) and \(D\) have coordinates \((8,0,0)\), \((0,6,0)\) and \((0,0,4)\) respectively. The point \(M\) is the midpoint of the edge \([EF]\).
(a)
Write down the coordinates of \(M\).
(1)
(b)
Find the length \(OM\).
(2)
(c)
Find the angle between the line \([OM]\) and the base \(OABC\).
(3)
(d)
Find the size of the angle \(M\hat{O}E\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(M(8,\,3,\,4)\)
A1
(b)
\(\sqrt{8^{2}+3^{2}+4^{2}}\)
(M1)
FT from their \(M\).
\(\sqrt{89}=9.43\) cm
A1
\(9.43398\ldots\)
(c)
identifying the right-angled triangle \(ONM\), where \(N(8,3,0)\) is vertically below \(M\)
(M1)
May be seen on the diagram.
\(\sin\theta=\dfrac{4}{\sqrt{89}}\) or \(\tan\theta=\dfrac{4}{\sqrt{73}}\)
(M1)
\(25.1^{\circ}\)
A1
\(25.0873\ldots^{\circ}\); \(0.438\) radians
(d)
\([ME]\) is perpendicular to the face \(OAED\), so triangle \(OEM\) has a right angle at \(E\)
(R1)
Or finding all three sides \(OE=\sqrt{80}\), \(EM=3\), \(OM=\sqrt{89}\) for the cosine rule.
\(\tan M\hat{O}E=\dfrac{3}{\sqrt{80}}\)
(M1)
Or \(\cos M\hat{O}E=\dfrac{80+89-9}{2\sqrt{80}\sqrt{89}}\).
78AASL3-q00123.1 Three-dimensional geometryPaper 2Hard9 marks
Diagram not to scale
The diagram shows a wedge \(ABCDEF\). The base \(ABCD\) is a horizontal rectangle with \(AB=12\) cm and \(BC=5\) cm. The face \(DCFE\) is a vertical rectangle with \(DE=CF=3\) cm. The sloping face \(ABFE\) is a rectangle, and its diagonals \([AF]\) and \([BE]\) are drawn.
(a)
Find the length \(AF\).
(2)
(b)
Find the angle between the line \([AF]\) and the base \(ABCD\).
(2)
(c)
Find the angle between the line \([AE]\) and the base \(ABCD\).
(2)
(d)
Find the acute angle between the diagonals \([AF]\) and \([BE]\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(AF=\sqrt{12^{2}+5^{2}+3^{2}}\)
(M1)
Or \(AC=13\) then \(\sqrt{13^{2}+3^{2}}\).
\(\sqrt{178}=13.3\) cm
A1
\(13.3416\ldots\)
(b)
\(F\) is vertically above \(C\): \(\sin F\hat{A}C=\dfrac{3}{\sqrt{178}}\) or \(\tan F\hat{A}C=\dfrac{3}{13}\)
(M1)
\(13.0^{\circ}\)
A1
\(12.9946\ldots^{\circ}\); \(0.227\) radians
(c)
\(E\) is vertically above \(D\): \(\tan E\hat{A}D=\dfrac{3}{5}\)
(M1)
\(31.0^{\circ}\)
A1
\(30.9637\ldots^{\circ}\); \(0.540\) radians
(d)
\(AE=\sqrt{34}\;(=5.83)\); the diagonals of rectangle \(ABFE\) bisect each other at \(X\)
(M1)
triangle \(AXE\) is isosceles; \(\tan\left(\tfrac{1}{2}A\hat{X}E\right)=\dfrac{\sqrt{34}/2}{6}\), so \(A\hat{X}E=2\arctan\dfrac{\sqrt{34}}{12}\)
(M1)
Or the cosine rule in triangle \(AXE\) with \(AX=XE=\dfrac{\sqrt{178}}{2}\).
\(51.8^{\circ}\)
A1
\(51.8314\ldots^{\circ}\); \(0.905\) radians. Award A0 for \(128^{\circ}\) only.
79AASL3-q00143.1 Three-dimensional geometryPaper 2Hard16 marks
Diagram not to scale
A bucket has the shape of a frustum of a right circular cone: a cone with its tip removed by a cut parallel to its base. The bottom of the bucket is a horizontal circle of radius \(5\) cm, the open top is a horizontal circle of radius \(10\) cm, and the depth of the bucket is \(12\) cm. When the sloping sides are extended they meet at the point \(V\), vertically below the centre of the bottom.
Ignore the thickness of the material.
(a)
Show that \(V\) is \(12\) cm below the bottom of the bucket.
(2)
(b)
Show that the volume of the bucket is \(700\pi\text{ cm}^{3}\).
(3)
(c)
Write down the capacity of the bucket in litres.
(1)
(d)
The outside of the bucket, including the bottom, is painted. Find the area that is painted.
(4)
(e)
The bucket contains \(1.5\) litres of water. Find the depth of the water.
(4)
(f)
Find the angle between a sloping side of the bucket and the horizontal.
(2)
Show mark scheme
Working
Mark
Notes
(a)
similar triangles: \(\dfrac{h}{5}=\dfrac{h+12}{10}\), where \(h\) is the distance of \(V\) below the bottom
80AASL3-q00223.2 Right-angled and non-right-angled trigonometryPaper 2Hard11 marks
In triangle \(ABC\), the point \(D\) lies on \([AC]\). It is given that \(AB=9\) cm, \(B\hat{A}D=65^{\circ}\), \(A\hat{B}D=52^{\circ}\) and \(DC=5\) cm.
Or the cosine rule in triangle \(ABC\) or \(BDC\). FT from (a), (b).
\(41.7^{\circ}\)
A1
\(41.6965\ldots^{\circ}\). The angle is acute because triangle \(BDC\) already contains the obtuse angle \(117^{\circ}\).
Answers: (a) \(9.15\) cm · (b) \(12.3\) cm · (c) \(52.9\text{ cm}^{2}\) · (d) \(41.7^{\circ}\)
81AASL3-q00283.3 Applications of trigonometryPaper 2Hard10 marks
The top of a lighthouse, \(L\), is \(60\) m vertically above sea level, and the base of the lighthouse is at \(O\). Two boats, \(A\) and \(B\), are on the sea. From \(L\), boat \(A\) is on a bearing of \(220^{\circ}\) with an angle of depression of \(8^{\circ}\), and boat \(B\) is on a bearing of \(295^{\circ}\) with an angle of depression of \(14^{\circ}\).
(a)
Find the horizontal distance \(OA\).
(2)
(b)
Find the horizontal distance \(OB\).
(1)
(c)
Find the distance between the two boats.
(3)
(d)
Find the bearing of \(B\) from \(A\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\tan8^{\circ}=\dfrac{60}{OA}\)
(M1)
Angle of elevation of L from A equals the angle of depression (alternate angles).
bearing of \(O\) from \(A\) is \(220^{\circ}-180^{\circ}=040^{\circ}\), and \(B\) lies anticlockwise of \([AO]\) as seen from \(A\)
(M1)
A correct diagram showing the north line at A is sufficient.
\(007.5^{\circ}\)
A1
\(7.48358\ldots^{\circ}\) Accept \(007^{\circ}\) or \(7.48^{\circ}\).
Answers: (a) \(427\) m · (b) \(241\) m · (c) \(432\) m · (d) \(007.5^{\circ}\)
82AASL3-q00313.3 Applications of trigonometryPaper 2Hard9 marks
A straight hillside slopes upwards at \(18^{\circ}\) to the horizontal. A vertical tree \([BT]\) grows on the hillside with its base at \(B\), where \(B\) is \(25\) m from the point \(P\) at the foot of the slope, measured along the slope. The point \(P\), the base \(B\) and the top of the tree \(T\) lie in the same vertical plane. The angle of elevation of \(T\) from \(P\) is \(48^{\circ}\).
(a)
Draw a clearly labelled diagram to represent this information.
(1)
(b)
Show that \(P\hat{B}T=108^{\circ}\).
(2)
(c)
Find the height of the tree, \(BT\).
(3)
(d)
Find the height of \(T\) above the horizontal through \(P\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
horizontal through \(P\); \([PB]\) at \(18^{\circ}\) labelled \(25\) m; vertical \([BT]\); \([PT]\) at \(48^{\circ}\) to the horizontal
A1
(b)
\([BP]\) makes an angle of \(18^{\circ}\) below the horizontal through \(B\), and \([BT]\) makes \(90^{\circ}\) above it
M1
Or: the slope makes \(90^{\circ}-18^{\circ}=72^{\circ}\) with the vertical on the uphill side, and \(P\hat{B}T=180^{\circ}-72^{\circ}\).
\(P\hat{B}T=90^{\circ}+18^{\circ}\)
A1
\(P\hat{B}T=108^{\circ}\)
AG
(c)
\(B\hat{P}T=48^{\circ}-18^{\circ}=30^{\circ}\) and \(P\hat{T}B=180^{\circ}-30^{\circ}-108^{\circ}=42^{\circ}\)
\(35.5332\ldots\) Or height of \(B\) above \(P\) \(=25\sin18^{\circ}=7.73\).
\(35.53\ldots\times\sin48^{\circ}\)
(M1)
Or \(7.725\ldots+18.68\ldots\).
\(26.4\) m
A1
\(26.4063\ldots\)
Answers: (c) \(18.7\) m · (d) \(26.4\) m
83AASL3-q00323.3 Applications of trigonometryPaper 2Hard14 marks
A yacht leaves port \(P\) and sails \(18\) km on a bearing of \(050^{\circ}\) to a buoy \(A\). It then sails \(25\) km on a bearing of \(130^{\circ}\) to a lighthouse \(B\).
(a)
Draw a clearly labelled diagram to represent this information.
(2)
(b)
Show that \(P\hat{A}B=100^{\circ}\).
(2)
(c)
Find the distance \(PB\).
(3)
(d)
Find the bearing of \(B\) from \(P\).
(3)
(e)
Write down the bearing of \(P\) from \(B\).
(1)
(f)
Find the shortest distance from \(A\) to the line \((PB)\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
north lines at \(P\) and \(A\); \([PA]\) labelled \(18\) km at \(50^{\circ}\) clockwise from north at \(P\)
A1
\([AB]\) labelled \(25\) km at \(130^{\circ}\) clockwise from north at \(A\), and \([PB]\) drawn
A1
(b)
bearing of \(P\) from \(A\) is \(050^{\circ}+180^{\circ}=230^{\circ}\)
M1
Or co-interior angles: angle between north at \(A\) and \([AP]\) is \(130^{\circ}\).
Accept \(098^{\circ}\). FT \(50^{\circ}+\) their angle.
(e)
\(277.8^{\circ}\)
A1
Accept \(278^{\circ}\). FT their (d) \(+180^{\circ}\).
(f)
area of triangle \(PAB=\dfrac{1}{2}(18)(25)\sin100^{\circ}\;(=222)\)
(M1)
Or \(d=18\sin A\hat{P}B\).
\(\dfrac{1}{2}(33.245\ldots)d=221.58\ldots\)
(M1)
Or \(18\sin47.778\ldots^{\circ}\).
\(13.3\) km
A1
\(13.3299\ldots\)
Answers: (c) \(33.2\) km · (d) \(097.8^{\circ}\) · (e) \(277.8^{\circ}\) · (f) \(13.3\) km
84AASL3-q00333.3 Applications of trigonometryPaper 2Hard16 marks
Diagram not to scale
The diagram shows a field \(ABCD\) in the shape of a quadrilateral. \(AB=85\) m, \(BC=60\) m, \(AD=95\) m, \(A\hat{B}C=108^{\circ}\) and \(C\hat{A}D=40^{\circ}\).
(a)
Find \(AC\).
(3)
(b)
Find the area of triangle \(ABC\).
(2)
(c)
Find \(CD\).
(3)
(d)
Find the area of the field.
(3)
(e)
A fence is to be built around the edge of the field. The fencing costs €18.50 per metre. Find the cost of the fence, giving your answer to the nearest euro.
(3)
(f)
A water pipe is to be laid in a straight line from \(C\) to the edge \([AD]\). Find the shortest possible length of the pipe.
Answers: (a) \(118\) m · (b) \(2430\text{ m}^{2}\) · (c) \(76.1\) m · (d) \(6030\text{ m}^{2}\) · (e) €5848 · (f) \(76.0\) m
85AASL3-q00343.3 Applications of trigonometryPaper 2Hard15 marks
Two fire-lookout towers, \(A\) and \(B\), stand on level ground, with \(B\) \(12\) km due east of \(A\). A fire is seen at point \(F\). From \(A\) the bearing of \(F\) is \(042^{\circ}\), and from \(B\) the bearing of \(F\) is \(316^{\circ}\).
(a)
Draw a clearly labelled diagram to represent this information.
(2)
(b)
Show that \(A\hat{F}B=86^{\circ}\).
(2)
(c)
Find the distance \(AF\).
(3)
(d)
Find the distance \(BF\).
(2)
(e)
A fire station \(C\) is \(5\) km due south of \(B\). Find the distance from \(C\) to the fire.
(3)
(f)
Find the bearing of \(F\) from \(C\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(A\) and \(B\) with \(B\) due east of \(A\), labelled \(12\) km, with north lines at \(A\) and \(B\)
A1
\(F\) north of \([AB]\), with the angles \(42^{\circ}\) at \(A\) and \(316^{\circ}\) (or \(44^{\circ}\) west of north) at \(B\) shown
A1
(b)
\(F\hat{A}B=90^{\circ}-42^{\circ}=48^{\circ}\) and \(F\hat{B}A=316^{\circ}-270^{\circ}=46^{\circ}\)
\(360^{\circ}-28.514\ldots^{\circ}=331.485\ldots^{\circ}\). Accept \(331^{\circ}\) or \(332^{\circ}\).
Answers: (c) \(8.65\) km · (d) \(8.94\) km · (e) \(13.0\) km · (f) \(331.5^{\circ}\)
86AASL3-q00433.4 Radian measure, arcs and sectorsPaper 2Hard10 marks
Diagram not to scale
Two circles have centres \(A\) and \(B\) and radii \(6\) cm and \(8\) cm respectively, with \(AB=10\) cm. The circles intersect at \(P\) and \(Q\). The shaded region is the region common to both circles.
(a)
Show that \(A\hat{P}B\) is a right angle.
(1)
(b)
Find \(P\hat{A}Q\), in radians.
(2)
(c)
Find \(P\hat{B}Q\), in radians.
(1)
(d)
Find the area of the shaded region.
(4)
(e)
Find the perimeter of the shaded region.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(AP^{2}+BP^{2}=36+64=100=AB^{2}\), so by the converse of Pythagoras’ theorem \(A\hat{P}B=\dfrac{\pi}{2}\)
A1
\(A\hat{P}B\) is a right angle
AG
(b)
\(\cos P\hat{A}B=\dfrac{6}{10}\) and \(P\hat{A}Q=2\,P\hat{A}B\)
(M1)
By symmetry about (AB).
\(1.85\)
A1
\(1.85459\ldots\)
(c)
\(2\arccos\dfrac{8}{10}=1.29\)
A1
\(1.28700\ldots\) Or \(\pi-1.85459\ldots\).
(d)
segment of circle \(A\): \(\dfrac{1}{2}(6)^{2}\left(1.854\ldots-\sin1.854\ldots\right)\)
(M1)
\(16.1\)
(A1)
\(16.1026\ldots\)
segment of circle \(B\): \(\dfrac{1}{2}(8)^{2}\left(1.287\ldots-\sin1.287\ldots\right)\;(=10.46)\)
(M1)
Or sector \(A\) \(+\) sector \(B\) \(-\) kite \(APBQ\): \(33.38+41.18-48\).
88AASL3-q00553.5 The unit circle and the ambiguous casePaper 2Hard15 marks
A lighthouse \(L\) is \(9\) km from a harbour \(H\), on a bearing of \(035^{\circ}\) from \(H\). A ship leaves \(H\) and sails due north at a constant speed of \(15\) km h−1. The light from the lighthouse can be seen clearly from the ship only when the ship is within \(6.5\) km of \(L\).
(a)
Find the shortest distance between the ship’s path and \(L\).
(2)
(b)
Let \(S\) be a position of the ship at which \(SL=6.5\) km. Show that \(\sin H\hat{S}L=0.794\), correct to three significant figures.
(2)
(c)
Find the two possible values of \(H\hat{S}L\).
(3)
(d)
Hence find the two possible distances \(HS\).
(4)
(e)
Find the length of time, in minutes, for which the light can be seen clearly from the ship.
(2)
(f)
Find the bearing of \(L\) from the ship at the moment the light first becomes clearly visible.
the light first becomes visible at \(HS=3.42\) km, where \(H\hat{S}L=127.4^{\circ}\); bearing \(=180^{\circ}-127.4^{\circ}\)
(M1)
\(052.6^{\circ}\)
A1
Accept \(053^{\circ}\).
Answers: (a) 5.16 km · (c) \(52.6^{\circ}\), \(127.4^{\circ}\) · (d) 3.42 km, 11.3 km · (e) 31.6 min · (f) \(052.6^{\circ}\)
89AASL3-q00783.7 Circular functions and their graphsPaper 2Hard16 marks
A Ferris wheel has radius \(20\) m and its centre is \(23\) m above the ground. The wheel rotates at a constant speed and completes one rotation every \(12\) minutes. A passenger boards at the lowest point of the wheel at time \(t=0\). The height of the passenger above the ground, \(h\) metres, \(t\) minutes after boarding, is modelled by
\(h(t)=a\cos(bt)+d\), where \(b\gt0\).
(a)
Write down the minimum height and the maximum height of the passenger above the ground.
(2)
(b)
Find the value of \(a\), the value of \(b\) and the value of \(d\).
(5)
(c)
Find the height of the passenger \(2.5\) minutes after boarding.
(2)
(d)
During one complete rotation, find the length of time for which the passenger is more than \(35\) m above the ground.
(4)
(e)
A second passenger boards at the lowest point exactly \(3\) minutes after the first passenger. Find the first time \(t\), where \(t\gt3\), at which the two passengers are at the same height above the ground, and find this height.
\(-20\cos\left(\dfrac{\pi t}{6}\right)+23=35\), or graph of \(h\) with \(y=35\)
(M1)
\(t=4.23\)
(A1)
\(4.22899\ldots\)
\(t=7.77\)
(A1)
\(7.77100\ldots\)
\(3.54\) minutes
A1
\(3.54200\ldots\)
(e)
\(h(t)=h(t-3)\)
(M1)
Second passenger’s height is \(h(t-3)\); GDC intersection.
\(t=7.5\) minutes
A1
\(37.1\) m
A1
\(37.1421\ldots\;(=23+10\sqrt{2})\)
Answers: (a) \(3\) m, \(43\) m · (b) \(a=-20,\ b=\dfrac{\pi}{6},\ d=23\) · (c) \(17.8\) m · (d) \(3.54\) minutes · (e) \(t=7.5\), \(37.1\) m
90AASL3-q00793.7 Circular functions and their graphsPaper 2Hard15 marks
Two buoys, \(A\) and \(B\), move up and down on the waves. Their heights, in metres, above the sea bed \(t\) seconds after a recording starts are modelled by
\(h_{A}(t)=1.4\sin(1.3t)+2.0\) and \(h_{B}(t)=0.9\cos(0.8t)+2.1\), for \(0\le t\le10\).
The graphs of \(h_{A}\) (solid) and \(h_{B}\) (dashed) are shown.
(a)
Write down the maximum height of buoy \(A\).
(1)
(b)
Find the period of \(h_{A}\).
(2)
(c)
Find the first time at which buoy \(A\) is at its maximum height.
(2)
(d)
Find the values of \(t\) for which the two buoys are at the same height.
(3)
(e)
Find the total time, during the 10 seconds, for which buoy \(A\) is higher than buoy \(B\).
(3)
(f)
Find the greatest vertical distance between the two buoys during the 10 seconds.
(2)
(g)
The recording continues beyond \(t=10\) and the model remains valid. Find the first time after \(t=10\) at which buoy \(B\) is at its maximum height.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(3.4\) m
A1
(b)
\(\dfrac{2\pi}{1.3}\)
(M1)
\(4.83\) s
A1
\(4.83321\ldots\)
(c)
\(1.3t=\dfrac{\pi}{2}\)
(M1)
Or use of GDC maximum.
\(t=1.21\) s
A1
\(1.20830\ldots\)
(d)
\(h_{A}(t)=h_{B}(t)\), e.g. intersections on GDC
(M1)
at least three correct values from \(0.548,\ 2.61,\ 4.41,\ 6.84,\ 9.75\)
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